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Find the curvature of r(t)=3ti+etj+e−tk at the point P(0,1,1). Enter an exact answer. Provide your answer below: κ=

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Answer:

To find the curvature of the curve defined by the vector-valued function r(t) = 3ti + etj + e^(-tk) at the point P(0, 1, 1), we'll need to compute several derivatives and then use the curvature formula:

κ = |r'(t) × r''(t)| / |r'(t)|^3

Let's start by finding the derivatives:

1. r(t) = 3ti + etj + e^(-tk)

2. r'(t) = (3i + ej - e^(-tk)k)

3. r''(t) = (0i + 0j + e^(-tk)k^2)

Now, plug these derivatives into the curvature formula:

r'(0) = 3i + ej - ek

r''(0) = 0i + 0j - 0k

| r'(0) × r''(0) | = |(3i + ej - ek) × (0i + 0j - 0k)| = |(3i + ej - ek) × 0k| = 0

|r'(0)|^3 = |3i + ej - ek|^3 = √(3^2 + 1^2 + 1^2)^3 = √11^3 = 11^(3/2)

Now, plug these values into the curvature formula:

κ = | r'(0) × r''(0) | / | r'(0) |^3

κ = 0 / (11^(3/2))

κ = 0

So, the curvature of the curve at the point P(0, 1, 1) is κ = 0.

Explanation:

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User Alxndr
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